Question 32
In a convex quadrilateral $ABCD$, the diagonals intersect at the point $E$ and $\angle AEB = 90^\circ$. A point $P$ is chosen on the side $[AD]$ other than $A$ so that $PE = EC$. The circumcircle of the triangle $BCD$ intersects the side $[AD]$ at the point $Q$ other than $A$. The circle passing through $A$ and tangent to the line $EP$ at $P$ intersects the line segment $[AC]$ at the point $R$. If $B, Q, R$ are collinear, then show that $\angle BCD = 90^\circ$.